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DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS1 Mark
MCQ
The value of the determinant $\begin{vmatrix}\text{x}&\text{x}+\text{y}&\text{x}+2\text{y}\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\\text{x}+\text{y}&\text{x}+2\text{y}&\text{x}\end{vmatrix}$ is:
  • A
    $9x^2(x + y)$
  • ✓
    $9y^2(x + y)$
  • C
    $3y^2(x + y)$
  • D
    $7x^2(x + y)$
✓

Answer

Correct option: B.
$9y^2(x + y)$
$\begin{vmatrix}\text{x}&\text{x}+\text{y}&\text{x}+2\text{y}\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\\text{x}+\text{y}&\text{x}+2\text{y}&\text{x}\end{vmatrix}$
$=\begin{vmatrix}-2\text{y}&\text{y}&\text{y}\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\-\text{y}&2\text{y}&-\text{y}\end{vmatrix} \ [$Applying $R_1 \rightarrow R_1 - R_2$ and $R_3\rightarrow R_3 - R_2]$
$=\text{y}^2\begin{vmatrix}-2&1&1\\\text{x}+2\text{y}&\text{x}&\text{x}+\text{y}\\-1&2&-1\end{vmatrix}\ [$Taking $(y)$ common from $R_1$ and from $R_3]$
$=\text{y}^2\begin{vmatrix}-2&-3&3\\\text{x}+2\text{y}&3\text{x}+4\text{y}&-\text{y}\\-1&0&0\end{vmatrix} \ [$Applying $C_2 \rightarrow C_2 + 2C_1$ and $C_3 \rightarrow C_3 - C_1]$
$=\text{y}^2\big[-1(3\text{y}-9\text{x}-12\text{y})\big]$
$=\text{y}^2[9\text{y}+9\text{x}]$
$=9\text{y}^2(\text{y}+\text{x})$
Hence, the correct option is $(b)$

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